Solved Special Triangles (45-45-90 and 30-60-90 ) Worksheet ... - Free Printable

Let's solve each of these special right triangle problems using the properties of 45-45-90 and 30-60-90 triangles.

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🔷 Special Triangle Ratios:

1. 45°-45°-90° Triangle
- Legs are equal: $ x $
- Hypotenuse: $ x\sqrt{2} $

2. 30°-60°-90° Triangle
- Side opposite 30°: $ x $
- Side opposite 60°: $ x\sqrt{3} $
- Hypotenuse (opposite 90°): $ 2x $

We'll use these ratios to find missing sides.

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✔ Problem 1: 45°-45°-90°

Given: Hypotenuse = $ 14\sqrt{3} $, angles 45°, 45°, 90°

In a 45-45-90 triangle:
- Hypotenuse = leg × $ \sqrt{2} $
- So: $ \text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{14\sqrt{3}}{\sqrt{2}} $

Simplify:
$$
\frac{14\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 14 \cdot \frac{\sqrt{6}}{2} = 7\sqrt{6}
$$

So both legs $ p $ and $ q $ are $ 7\sqrt{6} $

✔ $ p = 7\sqrt{6},\quad q = 7\sqrt{6} $

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✔ Problem 2: 30°-60°-90°

Given: Side opposite 30° = $ 6\sqrt{3} $, angle at top is 30°, so that side is shorter leg.

In 30-60-90:
- Short leg (opposite 30°) = $ x $
- Long leg (opposite 60°) = $ x\sqrt{3} $
- Hypotenuse = $ 2x $

Here, short leg = $ 6\sqrt{3} $ → So $ x = 6\sqrt{3} $

Then:
- Hypotenuse $ z = 2x = 2(6\sqrt{3}) = 12\sqrt{3} $
- Long leg $ y = x\sqrt{3} = 6\sqrt{3} \cdot \sqrt{3} = 6 \cdot 3 = 18 $

✔ $ z = 12\sqrt{3},\quad y = 18 $

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✔ Problem 3: 30°-60°-90°

Given: Side opposite 60° = $ \sqrt{5} $, angle at top is 60° → this is the long leg.

In 30-60-90:
- Long leg = $ x\sqrt{3} $
- So $ x\sqrt{3} = \sqrt{5} $ → $ x = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15}}{3} $

Now:
- Short leg $ u $ (opposite 30°) = $ x = \frac{\sqrt{15}}{3} $
- Hypotenuse $ v = 2x = 2 \cdot \frac{\sqrt{15}}{3} = \frac{2\sqrt{15}}{3} $

✔ $ u = \frac{\sqrt{15}}{3},\quad v = \frac{2\sqrt{15}}{3} $

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✔ Problem 4: 45°-45°-90°

Given: Hypotenuse = $ \frac{6\sqrt{2}}{5} $

Use: hypotenuse = leg × $ \sqrt{2} $

So:
$$
\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{6\sqrt{2}/5}{\sqrt{2}} = \frac{6}{5}
$$

So both legs $ r $ and $ s $ are $ \frac{6}{5} $

✔ $ r = \frac{6}{5},\quad s = \frac{6}{5} $

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✔ Problem 5: 45°-45°-90°

Given: Hypotenuse = $ \frac{13\sqrt{2}}{2} $

Again:
$$
\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{13\sqrt{2}/2}{\sqrt{2}} = \frac{13}{2}
$$

So both legs $ a $ and $ b $ are $ \frac{13}{2} $

✔ $ a = \frac{13}{2},\quad b = \frac{13}{2} $

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✔ Problem 6: 30°-60°-90°

Given: Side opposite 30° = 16 → this is the short leg = $ x $

So:
- Hypotenuse $ g = 2x = 32 $
- Long leg $ h = x\sqrt{3} = 16\sqrt{3} $

✔ $ g = 32,\quad h = 16\sqrt{3} $

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✔ Problem 7: 30°-60°-90°

Given: Hypotenuse = 18

In 30-60-90:
- Hypotenuse = $ 2x $
- So $ 2x = 18 $ → $ x = 9 $

Then:
- Short leg (opposite 30°) = $ x = 9 $ → this is $ k $
- Long leg (opposite 60°) = $ x\sqrt{3} = 9\sqrt{3} $ → this is $ j $

✔ $ j = 9\sqrt{3},\quad k = 9 $

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✔ Problem 8: 30°-60°-90°

Given: Side opposite 60° = 32 → this is the long leg = $ x\sqrt{3} $

So:
- $ x\sqrt{3} = 32 $ → $ x = \frac{32}{\sqrt{3}} = \frac{32\sqrt{3}}{3} $

Then:
- Short leg $ k $ (opposite 30°) = $ x = \frac{32\sqrt{3}}{3} $
- Hypotenuse $ L = 2x = 2 \cdot \frac{32\sqrt{3}}{3} = \frac{64\sqrt{3}}{3} $

✔ $ k = \frac{32\sqrt{3}}{3},\quad L = \frac{64\sqrt{3}}{3} $

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ p = 7\sqrt{6},\quad q = 7\sqrt{6} $ |
| 2 | $ z = 12\sqrt{3},\quad y = 18 $ |
| 3 | $ u = \frac{\sqrt{15}}{3},\quad v = \frac{2\sqrt{15}}{3} $ |
| 4 | $ r = \frac{6}{5},\quad s = \frac{6}{5} $ |
| 5 | $ a = \frac{13}{2},\quad b = \frac{13}{2} $ |
| 6 | $ g = 32,\quad h = 16\sqrt{3} $ |
| 7 | $ j = 9\sqrt{3},\quad k = 9 $ |
| 8 | $ k = \frac{32\sqrt{3}}{3},\quad L = \frac{64\sqrt{3}}{3} $ |

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